What is the arc length of #f(t)=(t-tsqrt(t-1),8-2t) # over #t in [1,4]#?

Answer 1

#7.26345#

we have #x(t)=t-tsqrt(t-1)# then #x'(t)=1-sqrt(t-1)-t/(2sqrt(t-1))#
#y'(t)=-2#

so we have to solve

#int_1^4sqrt((1-sqrt(t-1)-t/(2sqrt(t-1)))^2+4)dt#
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Answer 2

To find the arc length of the curve defined by the function ( f(t) = (t - t\sqrt{t - 1}, 8 - 2t) ) over the interval ( t ) in ([1,4]), you need to use the formula for arc length:

[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} , dt ]

Where ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) are the derivatives of the parametric equations representing the curve.

So, first, you need to find the derivatives:

[ \frac{dx}{dt} = 1 - \frac{1}{2\sqrt{t - 1}} - t\frac{1}{2\sqrt{t - 1}} ] [ \frac{dy}{dt} = -2 ]

Then, compute the integral:

[ L = \int_{1}^{4} \sqrt{\left(1 - \frac{1}{2\sqrt{t - 1}} - t\frac{1}{2\sqrt{t - 1}}\right)^2 + (-2)^2} , dt ]

Once you integrate this expression over the interval ([1,4]), you'll obtain the arc length of the curve.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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